We are bounded in
a nutshell of Infinite Space: Worksheet # 5, Problem #2: Let’s get into the
math of Stellar Temperatures
2. Blackbodies are
nice because they’re such simple objects. Their outward appearance is entirely
determined by their temperature. If there were cows in space, astronomers would
imagine them to be spherical blackbodies (and seriously, it wouldn’t be a bad
approximation). In this exercise we’ll take advantage of the relative
simplicity of blackbodies to derive some useful expressions that you’ll use
this term, and throughout tour astronomy career.
a. In astronomy, it is often
useful to deal with something called the “bolometric flux”, ot the energy per
area per time, independent of frequency. Integrate the blackbody flux \(F_\nu
(T)\) over all the frequencies to obtain
the bolometric flux emitted from a blackbody, \(F (T) \). You can do this using u substitution for
the variable \(u \equiv hv/kT\) .
This will allow you to split thing s into a temperature dependent term, and a
term comprising an integral over all frequencies. However, rather than solving
for the integral, just set all the constants to a new, single constant called \(\sigma\)
, which is also known as the Stefan-Boltzmann
constant, and retain the dependence on the temperature T. If you’re really into calculus, go ahead and show that \(\sigma
\approx 5.7 \times 10^{-5} erg~s^{-1}~cm^{-2}~K^{-4}\). Otherwise, commit this number to
memory.
b. The Wein Displacement Law:
Convert the units of the blackbody intensity from \( B_\nu (T) \) to \(B_\lambda (T) \) IMPORTANT: Remember that the amount of
energy in a frequency interval \(d\nu\) has to be exactly equal to the amount of energy in the corresponding
wavelength interval \(d\lambda\).
c. Derive the expression for
the wavelength \(\lambda_{max}\) corresponding to the peak of the intensity distribution at a given
temperature \(T\). (HINT: How do you
find the maximum of a function? Once you do this, again substitute \(u
\equiv h\nu / kT\) ). The expression you
end up with will be transcendental, but you can solve it easily to the first
order, which is good enough for this exercise.
d. The Rayleigh-Jeans Tail:
Next, let’s consider photon energies that are much smaller than the thermal
energy. Use a first-order Taylor expansion on the term \(e^{\frac{h\nu}{kT}}\)
to derive a simplified form of
\(B_\nu (T)\) in this low-energy regime.
(HINT: The Taylor Expansion of \(e^x \approx 1 + x\) .)
e. Write an expression for the
total power output of a blackbody with a radius R, starting with the expression
for \(F_\nu\). This
total energy output per unit time is also known as the bolometric luminosity,
L.
a) As we have seen in less detail at other points,
blackbodies are a good approximation for most objects in the universe. The most
“perfect” blackbody we have found is actually the CMB (Cosmic Microwave Background),
the afterglow of the Big Bang. Based on a relationship between the
frequency/wavelength of the observed light and the temperature of the light, we
find an equation which can relate these two, as well as be derived and
anti-derived to give broader descriptions.
From a previous problem, we know that a description
of the Flux of a blackbody, any object which absorbs and releases thermal
radiation, is represented by: \[ F_\nu (T) = \frac{2\pi}{c^2} \frac{\nu^3
h}{e^{h\nu/kT} -1} ,\] so if we integrate over all possible frequencies, going
to infinity, we have: \[ \int _0 ^\infty F_\nu (T) = \frac{2\pi h }{c^2} \int
_0 ^\infty \frac{\nu^3 }{e^{h\nu/kT} -1} d\nu ,\] where we can substitute \[ u
= \frac{h\nu}{kT} \] and \[u\frac{kT}{h} = \nu\] as well as \[du = \frac{h}{kT},
\] into \[ F(T) = \int _0 ^\infty \frac{\left(\frac{kT}{h}\right) u^3 }{e^{u}
-1} \left(\frac{kT}{h}\right)du\] which can be more easily integrated once the
constants are completely separated: \[ F(T) = \frac{2\pi h }{c^2}
\left(\frac{kT}{h}\right)^4 \int _0 ^\infty \frac{ u^3 }{e^{u} -1} du,\] and
knowing a particular integration of a strange equation, this all becomes much
easier \[\int _0 ^\infty \frac{ u^3 }{e^{u} -1} du = \frac{\pi^4}{15}\] \[ F(T)
= \frac{2\pi k^4 }{c^2 h^3} T^4 \cdot \frac{\pi^4}{15}.\] As is stated in the
problem, all the constants can be grouped together and you would find that we
come remarkably close to the actual value of the Stefan-Boltzmann constant, although I suspect
we would get a closer value were we to complete the integration by ourselves,
as difficult as that may be. Nevertheless, the constants all become a single
term, which simplifies the Bolometric Flux equation into: \[ F(T)= \sigma T^4\]
b) From the problem, we know we are simply
converting the specific intensity according to frequency, to specific intensity
according to wavelength. Therefore, if they are both the same value, but in
different metrics, they can simply be equaled to the other and then attempt to
integrate both, with parameters which leave us with their derivatives with
respect to the parameter: \[ -B_\lambda d\lambda = B_\nu d\nu ,\] (we add a negative sign to establish how they are inversely proportional) and by placing
all on one side: \[ -B_\lambda = -B_\nu \frac{d\nu}{d\lambda} ,\] and after simplifying it down: \[ B_\lambda = B_\nu \frac{d\nu}{d\lambda} ,\] we just find the derivative of \( \nu
= \frac{c}{\lambda}\) in terms of \(\lambda\): \[ B_\lambda = B_\nu~ \frac{c}{\lambda^2},\]
and so applying it to the original intensity equation (remember to change all \(\nu\)’s
into \(c/\lambda\)) : \[ B_\lambda = \frac{2hc^2}{\lambda^5} \frac{1}{e^{hc/\lambda
kT} – 1} .\]
c) Now, to maximize the equation, we have to derive
the function and set it equal to ): \[ \frac{d
~B_\lambda}{d\lambda} = \frac{d \frac{2hc^2}{\lambda^5} \frac{1}{e^{hc/\lambda
kT} – 1}}{d\lambda} ,\] which can now be interpreted with partial fractions
(only deriving with respect to a specific variable: \[\frac{\partial B_\lambda}{\partial
d\lambda} = \frac{\partial B\lambda}{\partial u} \frac{\partial u}{\partial \lambda}\]
and thus we can define u as \[ u = \frac{hc}{kT\lambda},\] and we place thisn
into the equation and get: \[ B_u = 2hc^2
\frac{\left(\frac{ukT}{hc}\right)^5}{e^u – 1} ,\] and now we take the derivative
with respect to u on both sides and get (with
some separation of constants and application of the product rule): \[ \frac{d
B_u}{du} = 2hc^2 \left(\frac{kt}{hc}\right)^5 (5u^4(e^u -1 ) – u^5(e^u)),\] so we now take
it and optimize it by setting it to 0: \[ 0 = 2hc^2 \left(\frac{kt}{hc}\right)^5
(5u^4(e^u -1 ) – u^5(e^u)),\] we
eliminate all the constants and begin solving the equation: \[ 5u^4 (e^u -1 ) =
u^5 ~e^u \] \[ 5 = \frac{ue^u}{e^u -1 },\] and now we can make a first order
expansion and simplify the equation: \[ 5 = \frac{u e^u}{e^u} \] \[ 5 = u\]
Knowing what we defined to be u: \[ u = \frac{hc}{kT\lambda}\]
and solved to find \(\lambda\), we see: \[ \lambda = \frac{hc}{ukT},\] and now
we define our constants, the Plank constant: \[ h =6.63 \times 10^{-27} erg \cdot
s,\] Boltzmann’s constant: \[ k = 1.38 \times 10^{-16} erg/K,\] and the speed
of light: \[ c = 3 \times 10^{10} cm/s,\]
we solve and find that: \[ \lambda_{max}
= \frac{6.63 \times 10^{-27} erg \cdot s \cdot 3 \times 10^{10} cm/s }{ 5 \cdot
1.38 \times 10^{-16} erg/K \cdot T} \]
\[ \lambda_{max} = \frac{0.29 cm \cdot K}{T }\]
d) This part of the problem is simply using a basic
equation and expanding it to see how it acts in small frequencies, \[B_\nu(T) =
\frac{2\nu^3 h }{c^2 (e^{h\nu / kT} -1)}\] and the expansion is given to us as:
\[e^x \approx 1 + x\] which, plugged in, would just be: \[ B_\nu(T) = \frac{2\nu^3 h }{c^2} \frac{1}{1
+ \frac{h\nu}{ kT} -1},\] and thus we are left with: \[ B_\nu(T) = \frac{2\nu^3
h }{c^2} \frac{1}{\frac{h\nu}{ kT}}\]
e) Now, just applying what we learned in part a),
we know how\[ F_\nu (T) = \frac{2\pi}{c^2} \frac{\nu^3 h}{e^{h\nu/kT} -1} ,\]
can be integrated over all frequencies and we find: \[ F(T) = \sigma T^4,\]then,
we have to understand how Flux is distributed isotopically. This means we have
to understand the entire area, this being: \[ A = \int^{2\pi}_0 \int^\pi_0 R^2
\sin\theta d\theta d\phi\] \[ A= \int^{2\pi}_0
2R^2 d\phi\] \[ A = 4\pi R^2 \] So by multiplying both sides of the \(F(T)\)
equation by the complete area of the light expansion, we have: \[ 4\pi R^2 F =
4\pi R^2 \sigma T^4,\] and Flux in all directions at all times is simply the
Luminosity, as such: \[ L = 4\pi R^2 \sigma T^4,\] the relationship between
Luminosity and Temperature.
